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BASIC INSTRUMENTATION ELECTRICITY. 1.5V. Voltage and Current. R. R. I. Q=P/R. I=V/R. Liquid flow. Electrical current. Resistance. Every substance has resistance Conductor is substance having low resistance Isolator is substance having high resistance

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voltage and current

1.5V

Voltage and Current

R

R

I

Q=P/R

I=V/R

Liquid flow

Electrical current

resistance
Resistance
  • Every substance has resistance
  • Conductor is substance having low resistance
  • Isolator is substance having high resistance
  • 16 AWG wire resistance is ±12 Ω/km
  • 18 AWG wire resistance is ± 20 Ω/km
  • Question:
    • What is the resistance of 600 m 16 AWG wire?
voltage drop
Voltage Drop
  • When current flows across a wire the voltage will drop
  • Example

I=16 mA

PT

V=24 V

length= 500 m

What is the voltage across the PT

problem
Problem
  • The allowed voltage for a Pressure Transmitter is 18V to 30 V. What is the maximum wire length if the power supply voltage in the control room is 24 V?
ac voltage and current

VAC

R

Vm

AC VOLTAGE AND CURRENT

v(t)

0

ωt

v(t) = Vmcos(ωt + )

Frequency , f = 50 Hz/ 60 Hz

T = 1/f = 1/50 = 0.02 s

ω = 2πf

 is the phase angle

ac voltage and current in resistor

v(t)

VAC

R

Vm

Im

ωt

i(t)

ωt

V and I in resistor are in phase

AC VOLTAGE and CURRENT in RESISTOR

v(t) = Vmcos ωt

i(t) = v(t)/R

= Vm (cos ωt)/R

= Imcosωt

Im = Vm/R

ac voltage current and power in resistor

v(t)

ωt

i(t)

AC VOLTAGE, CURRENT and POWER in RESISTOR

v(t) = Vmcosωt

i(t) = Imcosωt

p(t) =v(t) i(t)

= VmImcos2ωt

= VmIm {1+cos(2ωt )}/2

p(t)

capacitor

+

+

+

+

-

-

-

-

+ -

+ -

CAPACITOR

Unit of C is F (Farad)

1 Farad = 1 Coul/Volt = 1As/V

Real capacitor always have intrinsic capacitor and resistor with it

voltage and current in capacitor

Im

Vm

v(t)

i(t)

ωt

0

π/2

VOLTAGE AND CURRENT IN CAPACITOR

v(t) = Vmcos(ωt +)

VAC

C

The current lead the voltage

power in capacitor

VAC

C

p(t)

POWER IN CAPACITOR

v(t)

i(t)

v(t) = Vmcos(ωt +)

+

+

p(t) = v(t) i(t)

= VmImcos(ωt + )sin(ωt + )

-

-

inductor

i(t)

coil

core

inductor

Equivalent Ckt of inductor

inductor

Unit of L is H(Henry)

1 H = 1 Vs/A

Inductor is made of coil and core.

Real inductors always have intrinsic capacitor and resistor with it

voltage and current in inductor

v(t)

i(t)

v(t)

Vm

Im

ωt

VOLTAGE AND CURRENT IN INDUCTOR

i(t)

for

The voltage lead the current

Wehave

power in inductor

v(t)

p(t)

POWER IN INDUCTOR

v(t)

i(t)

v(t) =  Vmsin(ωt +)

+

+

-

p(t) = v(t) i(t)

= VmImcos(ωt + )sin(ωt + )

-

power in rl circuit

v(t)

i(t)

v(t)

i(t)

ωt

0

p(t)

+

+

-

-

Power in RL circuit
voltage and current rc circuit

v(t)

v(t)

C

i(t)

i(t)

ωt

for

VOLTAGE AND CURRENT RCCIRCUIT
power in rc circuit

v(t)

C

i(t)

v(t)

i(t)

ωt

0

p(t)

+

+

-

-

Power in RC circuit
ac voltage current and power in r l and c summery
AC VOLTAGE, CURRENT and POWER in R, L, and C (summery )

v

v

v

i

i

i

ωt

ωt

ωt

p(t)

RESISTOR

CAPACITOR

INDUCTOR

power in rc circuit1

v(t)

v(t)

i(t)

i(t)

ωt

0

ωt

0

p(t)

+

+

p(t)

+

+

-

-

-

-

Power in RC circuit

RL CIRCUIT

RC CIRCUIT

phasors
Phasors
  • A phasor is a complex number that represents the magnitude and phase of a sinusoid:
example for v and i phasor in resistor

VAC

R

Vm/√2

v(t)

Im /√2

i(t)

ωt

Example for V and I phasor in resistor

v(t) = Vmcos(ωt + )

i(t) = Vm/Rcos(ωt + )

example for v and i phasor in capacitor

VAC

C

v(t)

i(t)

Vm

Im

ωt

0

Im /√2

Vm/√2

Example for V and I phasor in capacitor

v(t) = Vmcos (ωt+)

example for v and i phasor in capacitor1

Im/√2

Vm/√2

Example for V and I phasor in capacitor

Im/√2

Vm/√2

We can set the angle  arbitrarily. Usually we set the voltage is set to be zero phase abritrary

v(t) = Vmcos ωt

example for v and i phasor in inductor

Vm/√2

Im/√2

Vm/√2

Im/√2

Example for V and I phasor in inductor

v(t)

i(t)

Here we can set the voltage to be zero phase, then the phase of current will be 

impedance
Impedance
  • By definition impedance (Z) is

Z = V/I

  • AC steady-state analysis using phasors allows us to express the relationship between current and voltage using a formula that looks likes Ohm’s law:

V = IZ

impedance cont d
Impedance (cont’d)
  • Impedance depends on the frequency w.
  • Impedance is (often) a complex number.
  • Impedance is not a phasor (why?).
  • Impedance allows us to use the same solution techniques for AC steady state as we use for DC steady state.
  • Impedance in series/parallel can be combined as resistors
impedance of resistor

VAC

R

Vm/√2

Im /√2

Impedance of resistor

v(t) = Vmcos(ωt + )

i(t) = Vm/Rcos(ωt + )

ZR = R

impedance of capacitor

VAC

C

Im /√2

Vm/√2

Impedance of capacitor

v(t) = Vmcos (ωt+)

impedance of capacitor inductor
Impedance of capacitor inductor

ωLIm/√2

v(t)

i(t)

Im/√2

ZL = jωL

impedance1
Impedance

ZR = R

ZL = jωL

impedance example

+

1mF

-

Impedance Example:

f = 50Hz

Find ZC

Answer:

Zc = 1/jwC

w=2pf =2 × 3.14 × 50 = 314 rad/s

Zc = 1/jwC=1/(j × 314 × 106)

Zc = j3184.71

symbol of impedance

Z

ZT = Z1+ Z2

Z1

Z2

Z1

Z2

Symbol of Impedance

Impedance in series

Impedance in parallel

impedance in series example
Impedance in series example

R = 1K2

C = 15 mF

w = 314

ZT = ?

Answer:

Zc = 1/jwC=1/(j × 314 × 15 × 106)

Zc = j212.31

ZT = 1200 – j212.31

impedance in series example1
Impedance in series example

R = 1K2

ZT = ?

L = 5 mH

w = 314

Answer:

ZL = jwL=j × 314 × 5 × 103

ZL = j1.57

ZT = 1200 + j1.57

impedance in series example2

R = 1K2

C = 15 mF

w = 314

Zc = j212.31

ZT = ?

L = 5 mH

w = 314

ZL = j1.57

Impedance in series example

Answer:

ZT = 1200 – j212.31 + j1.57

= 1200 –j210.74

impedance resistance and reactance
Impedance, Resistance, and Reactance

Generally impedance consist of:

The real part which is called Resistance, and

The imaginary part which is called reactance

Z = R + jX

impedance

Resistance

reactance

example single loop circuit

20kW

+

+

VC

10V  0

1mF

-

-

Example: Single Loop Circuit

w = 377 Find VC

example cont d
Example (cont’d)
  • How do we find VC?
  • First compute impedances for resistor and capacitor:

ZR = 20kW= 20kW  0

ZC = 1/j (377 1mF) = 2.65kW  -90

impedance example cont d

20kW  0

+

+

VC

2.65kW  -90

Vi =10V  0

-

-

Impedance Example (cont’d)

Then use the voltage divider to find VC:

impedance example cont d1

+

-

Impedance Example (cont’d)

20kW  0

+

VC

2.65kW  -90

Vi =10V  0

-

complex power
Complex Power

Complex power is defined as

S = VI*

The unit of complex power is Volt Ampere (VA)

S= VI* = I2Z = I2(R+jX) = I2R+jI2X

= I2Z cos  +jI2Z sin 

S = VI cos  +jVI sin  = P + jQ

Sis called apparent power and the unit is va

Pis called active power and the unit is watt and

Qis called reactive power and the unit is var

signal conditioner
SIGNAL CONDITIONER
  • Signals from sensors do not usually have suitable characteristics for display, recording, transmission, or further processing.
  • They may lack the amplitude, power, level, or bandwidth required, or they may carry superimposed interference that masks the desired information.
signal conditioner1
SIGNAL CONDITIONER
  • Signal conditioners, including amplifiers, adapt sensor signals to the requirements of the receiver (circuit or equipment) to which they are to be connected.
  • The functions to be performed by the signal conditioner derive from the nature of both the signal and the receiver. Commonly, the receiver requires a single-ended, low-frequency (dc) voltage with low output impedance and amplitude range close to its power-supply voltage(s).
signal conditioner2
SIGNAL CONDITIONER
  • A typical receiver here is an analog-to-digital converter (ADC).
  • Signals from sensors can be analog or digital. Digital signals come from position encoders, switches, or oscillator-based sensors connected to frequency counters.
  • The amplitude for digital signals must be compatible with logic levels for the digital receiver, and their edges must be fast enough to prevent any false triggering.
  • Large voltages can be attenuated by a voltage divider and slow edges can be accelerated by a Schmitt trigger.
operational amplifiers
Operational Amplifiers

The term operational amplifier or "op-amp" refers to a class of high-gain DC coupled amplifiers with two inputs and a single output. The modern integrated circuit version is typified by the famous 741 op-amp. Some of the general characteristics of the IC version are:

  • High input impedance, low output impedance
  • High gain, on the order of a million
  • Used with split supply, usually +/- 15V
  • Used with feedback, with gain determined by the feedback network.
non inverting amplifier
Non-inverting Amplifier

For an ideal op-amp, the non-inverting amplifier gain is given by

voltage follower
Voltage Follower

The voltage follower with an ideal op amp gives simply

But this turns out to be a very useful service, because the input impedance of the op amp is very high, giving effective isolation of the output from the signal source. You draw very little power from the signal source, avoiding "loading" effects. This circuit is a useful first stage.

The voltage follower is often used for the construction of buffer for logic circuits.

current to voltage amplifier
Current to Voltage Amplifier

A circuit for converting small current signals (>0.01 microamps) to a more easily measured proportional voltage.                                             

so the output voltage is given by the expression above.

voltage to current amp
Voltage-to-Current Amp

The current output through the load resistor is proportional to the input voltage

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